Solution:
If I look at the dot-product of [tex]\frac{d}{dt}(\vec{\theta} \cdot \vec{u}}) = \vec{\theta(t)^{\prime}} \cdot \vec{u(t)}[/tex]

Next by Fundamental theorem of Calculus leftside of (1) is true.

The inequality however is a bit tricky for me. I seem to remember an inequality
which states that the dot product of two vectors a and b can be written as
[tex]\vec{a} \cdot \vec{b} \leq |\vec{a}| \cdot |\vec{b}|[/tex] which transfered to Our situation gives
[tex]\vec{\theta'(t)} \cdot \vec{u(t)} \leq |\vec{\theta'(t)}| \cdot |\vec{u(t)}|[/tex] and since [tex]|\vec{u}| = 1[/tex], then this is equivalent to [tex]\theta'(t) \cdot \vec{u(t)} \leq |\theta'(t)| [/tex] and finally by the rules regarding our Riemann Integrals [tex]\int_{a}^{b} f \leq \int_{a}^{b} g[/tex], then the integrals [tex]\int_{a}^{b} \ldots dt \leq \int_{a}^{b} \ldots \ dt[/tex] is also true.

There is a second part as well

If I set [tex]\vec{u} = \frac{l-m}{|l-m|}[/tex]

Then show that [tex]|\theta(b) - \theta(a)| \leq \int_{a}^{b} |\vec{\theta(t)^{\prime}| dt[/tex]

If I insert [tex]\vec{u}[/tex] into lefthand side of the inequality I get [tex](l-m) \cdot \frac{l-m}{|l-m|} = |l-m|[/tex] since [tex](l-m) \cdot (l-m) = |l-m|^2[/tex]

From this it should follow that the length of the curve from l to m is a straight line joining these points. But how exactly does it do that?
Is it something to do with if I on my curve draw a straight line from l to m and use this line as a side in a triangle and by using pythagoras then claim that line previously mentioned always will be shorter than the arc length from l to m?